Strength of Materials
Strength of materials (mechanics of materials) connects external loads to internal stresses and deformations. It is the bridge from statics to real design: given a load, will the part yield, deflect too far, twist, or buckle? The exam concentrates on axial members, beams in bending, shafts in torsion, and the factor of safety.
Stress, strain, and Hooke's law
Normal stress is force per unit area, σ = P/A; a 10,000 N load on 0.002 m² gives σ = 5 MPa. Shear stress is τ = V/A, force acting parallel to the area. Normal strain is the fractional length change ε = δ/L (dimensionless); a 2 m bar stretching 2 mm has ε = 0.001. In the elastic region, Hooke's law ties them together through the modulus of elasticity: σ = E·ε. A material with E = 200 GPa at ε = 0.001 carries σ = 200 MPa. Watch units: 1 GPa = 10⁹ Pa, and 1 MPa = 10⁶ Pa.
Axial deformation
An axially loaded prismatic bar elongates by δ = P·L/(A·E). The deformation grows with load and length and shrinks with cross-sectional area and stiffness. A steel bar (E = 200 GPa), 2 m long, area 1×10⁻⁴ m², under 10,000 N stretches δ = (10,000·2)/(1×10⁻⁴·200×10⁹) = 1×10⁻⁴ m = 0.1 mm. This single formula also underlies thermal-stress and statically indeterminate problems, where compatibility of deformations supplies the extra equation beyond equilibrium.
Beams: shear, moment, and bending stress
Transverse loads create internal shear V and bending moment M that vary along the beam; shear and moment diagrams track them. For a simply supported span L with a central point load P, the maximum moment is M = P·L/4 — an 8 kN load on a 4 m span gives M = 8·4/4 = 8 kN·m. For a uniformly distributed load w, the maximum moment is w·L²/8. Bending stress is σ = M·c/I, largest at the outer fiber (distance c from the neutral axis), where I is the area moment of inertia of the cross-section. This is why deep sections carry bending far better than shallow ones.
Torsion of circular shafts
A torque T twisting a circular shaft produces shear stress τ = T·r/J, maximum at the outer surface (r), where J is the polar moment of inertia — J = π·D⁴/32 for a solid shaft. A 0.05 m diameter shaft carrying 1000 N·m has J = π(0.05)⁴/32 ≈ 6.14×10⁻⁷ m⁴ and surface stress τ = 1000·0.025/6.14×10⁻⁷ ≈ 40.7 MPa. The angle of twist is φ = T·L/(J·G), where G is the shear modulus. Confusing J (polar) with I (rectangular bending inertia) is a frequent slip.
Factor of safety and failure criteria
The factor of safety compares a material's strength to the stress it actually sees: FS = σ_yield/σ_working. A part with yield strength 250 MPa at a working stress of 100 MPa has FS = 2.5. Design codes prescribe minimum values. For ductile materials, yielding is predicted by the maximum-shear (Tresca) or distortion-energy (von Mises) criterion under combined stress; brittle materials use maximum-normal-stress. Higher FS buys margin against overload, material variability, and uncertainty in the analysis, at the cost of weight and material.
Column buckling and combined loading
Slender columns in compression can fail by buckling — a sudden sideways deflection — well below the yield stress. The Euler critical load is P_cr = π²·E·I/(K·L)², where K depends on the end conditions (K = 1 pinned–pinned, 0.5 fixed–fixed, 2 fixed–free). Buckling capacity rises with stiffness E·I and falls with the square of length, so long slender members are most at risk. When axial, bending, and torsional loads act together, superposition adds their stresses at a point before a failure criterion is applied.
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